Page 71 - Advanced engineering mathematics

2.2 The Constant Coefficient Case 51

Consider the constant-coefficient linear homogeneous equation

y + ay + by = 0 (2.3)

in which a and b are constants (numbers). A method suggests itself if we read the differential
equation like a sentence. We want a function y such that the second derivative, plus a constant
multiple of the first derivative, plus a constant multiple of the function itself is equal to zero for
λx
λx
all x. This behavior suggests an exponential function e , because derivatives of e are constant
λx
λx
multiples of e . We therefore try to find λ so that e is a solution.
λx
Substitute e into equation (2.3) to get
2 λx
λx
λx
λ e + aλe + be = 0.
Since e λx is never zero, the exponential factor cancels, and we are left with a quadratic
equation for λ:
2
λ + aλ + b = 0. (2.4)
The quadratic equation (2.4) is the characteristic equation of the differential equation (2.3).
Notice that the characteristic equation can be read directly from the coefficients of the differential
equation, and we need not substitute e each time. The characteristic equation has roots
λx
1 √
2
(−a ± a − 4b),
2
leading to the following three cases.
Case 1: Real, Distinct Roots
This occurs when a − 4b > 0. The distinct roots are
2
1 √ 1 √
2
2
λ 1 = (−a + a − 4b) and λ 2 = (−a − a − 4b).
2 2
e λ 1 x and e λ 2 x are linearly independent solutions, and in this case, the general solution of
equation (2.3) is
y = c 1 e λ 1 x + c 2 e λ 2 x .

EXAMPLE 2.6
From the differential equation

y − y − 6y = 0,
we immediately read the characteristic equation
2
λ − λ − 6 = 0
as having real, distinct roots 3 and −2. The general solution is
3x
y = c 1 e + c 2 e −2x .

Case 2: Repeated Roots
2
This occurs when a − 4b = 0 and the root of the characteristic equation is λ =−a/2. One
solution of the differential equation is e −ax/2 .
We need a second, linearly independent solution. We will invoke a method called reduction
of order, which will produce a second solution if we already have one solution. Attempt a second
solution y(x) = u(x)e −ax/2 . Compute
a
−ax/2
y = u e − ue −ax/2

2

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October 14, 2010 14:12 THM/NEIL Page-51 27410_02_ch02_p43-76